Operators §

Bounded is the same as continuous §

For a linear map, the two notions coincide. If $T$ is continuous at $0$, then $\|Tx\| \leq M\|x\|$ for small $x$, and scaling gives it for all $x$. Conversely, the estimate $\|Tx - Tx_0\| = \|T(x - x_0)\| \leq M\|x - x_0\|$ shows continuity everywhere. So “bounded” is just “continuous” wearing a linear hat.

Example. On $\ell^2$, the diagonal map $T(x_1, x_2, \dots) = (x_1, x_2/2, x_3/3, \dots)$ has $\|T\| = 1$ (attained on the first basis vector). The shift $S(x_1, x_2, \dots) = (0, x_1, x_2, \dots)$ is an isometry, so $\|S\| = 1$.

The space of operators is itself a Banach space §

The set $\mathcal{B}(X, Y)$ of all bounded operators $X \to Y$, with the operator norm, is a Banach space whenever $Y$ is. This is important: it means you can take limits of operators, apply fixed-point theorems to them, and treat “operators” as first-class points of a space.

Why operators must sometimes be unbounded §

Many natural operations are not bounded. Differentiation is the prime example: on $L^2([0,1])$, the map $A f = f'$ is unbounded, because a sequence of functions can have arbitrarily small $L^2$ size while their derivatives do not. Yet differentiation is central to physics and PDEs, so functional analysis needs a home for it: the densely defined, unbounded operator.

Example. The momentum operator in quantum mechanics is $A = -i\hbar\, d/dx$ on a suitable domain in $L^2(\mathbb{R})$. It is unbounded but densely defined, and (on the right domain) self-adjoint — which is exactly what makes its spectrum (the possible measured momenta) real.

The domain is part of the operator §

Two operators with the same formula but different domains are different operators. This is why “densely defined” matters: it is the condition that lets the adjoint (below) exist.

The adjoint turns the inner product around §

The adjoint is the operator that lets you “move $T$ across the inner product.” In finite dimensions this is just the conjugate transpose of a matrix. The identity $\langle Tx, y \rangle = \langle x, T^* y \rangle$ is the whole definition; everything else follows from it.

Example. If $T$ is the matrix $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then $T^* = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. If $T$ is multiplication by a function $g$ on $L^2$, then $T^*$ is multiplication by $\overline{g}$.

Self-adjoint means real spectrum §

A self-adjoint operator has only real eigenvalues: if $T x = \lambda x$ with $\|x\| = 1$, then $\lambda = \langle Tx, x \rangle = \langle x, T^* x \rangle = \overline{\langle Tx, x \rangle} = \overline{\lambda}$, so $\lambda \in \mathbb{R}$. This is why observables in quantum mechanics are modeled by self-adjoint operators — their measured values are real.

Compact means “almost finite rank” §

In finite dimensions every operator is compact. In infinite dimensions, compact operators are the ones that behave like finite-rank operators: they squeeze an infinite-dimensional ball into something “small” (with compact closure). The key structural fact is that every compact operator on a Banach space is the norm-limit of finite-rank operators, so compact operators are the infinite-dimensional shadow of matrices.

Examples.

• Every finite-rank operator is compact.
• An integral operator $(Tf)(x) = \int K(x, y) f(y)\,dy$ with a square-integrable kernel $K$ (a Hilbert–Schmidt operator) is compact on $L^2$.

Why compactness matters §

The spectrum of a compact operator is “discrete”: except possibly for $0$, it consists of eigenvalues of finite multiplicity that accumulate only at $0$. This is what makes the eigenvalue expansions of Fourier theory and the spectral theorem for compact operators so clean, and it is the reason integral equations can be solved by expanding in eigenfunctions.

From diagonalizing a matrix §

In finite dimensions, a normal matrix is unitarily diagonalizable: $T = U \Lambda U^*$ with $\Lambda$ diagonal. The spectral theorem is the infinite-dimensional version of exactly this statement. The “diagonal entries” become a function $f$, the “basis of eigenvectors” becomes a measure space, and “diagonalizing” becomes “multiplication by $f$.”

The projection-valued measure §

The precise form: to a self-adjoint $T$ there is a unique projection-valued measure $E$ on the Borel sets of $\mathbb{R}$ such that

The spectrum of $T$ is the support of $E$. This formula is the infinite-dimensional replacement for $T = \sum \lambda_i P_i$ (eigenvalues times projections).

Consequences §

• Real spectrum: a self-adjoint operator has real spectrum (the integral is over $\mathbb{R}$).
• Functional calculus: for any Borel function $g$, you can define $g(T) = \int g(\lambda)\, dE(\lambda)$. This lets you write $e^{i t T}$ (the time-evolution operator in quantum mechanics) and $\sqrt{T}$.
• Eigenfunction expansions: when the spectrum is discrete (compact case), the formula reduces to an expansion in eigenfunctions, as in Fourier series.

Problem. Let $T$ be a compact self-adjoint operator on a Hilbert space with $0$ not in its spectrum. Show that $T$ is invertible and that $T^{-1}$ is bounded.